SRM 573 DIV2

1、照旧水题~~~，略过

2、贪心算法

#include <iostream> 
#include <string> 
#include <vector> 
#include <cstdlib> 
#include <cmath> 
#include <map> 
#include <stack> 
#include <algorithm> 
#include <list> 
#include <ctime> 
#include <set> 
#include <queue> 
typedef long long ll; 
using namespace std; 

class TeamContestEasy { 
public: 
  int worstRank(vector <int> str){ 
    int sz=str.size(); 
    int point=str[1]+str[2]+str[0]-min(str[0],min(str[1],str[2])); 
    if(3==sz) 
      return 1; 
    vector<int> part(sz-3,0); 
    copy(str.begin()+3,str.end(),part.begin()); 
    sort(part.begin(),part.end()); 
    reverse(part.begin(),part.end()); 
    int sz2=part.size()-1; 
    int szo=part.size()/3; 
    int start=0; 
    int res=0; 
/* 
    for(int i=0;i<=sz2;i++) 
      cout<<part[i]<<endl; 
*/ 
    while(start<sz2){ 
      if((part[start]+part[sz2])>point){ 
        res++; 
        start++; 
        sz2--; 
      }else{ 
        sz2--; 
      } 
    } 
    if(res>szo) 
      res=szo; 
    return (res+1); 

  } 
};

3、DP的算法，简化一下，先求一个点Ai到达矩形内某个点Bi的路径数目，然后相乘就是到达所有点A到达点Bi的数目，将点Bi的数目相加就是总数目~~~

#include <iostream>
#include <string>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <map>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime>
#include <set>
#include <queue>
typedef long long ll;
using namespace std;

class WolfPackDivTwo {
public:
    int calc(vector<int>, vector<int>, int);
};

#define maxnum 160
#define MOD 1000000007
int res[maxnum][maxnum];
int dp[2][maxnum][maxnum];
const int dx[] = { 0, 0, 1, -1 };
const int dy[] = { 1, -1, 0, 0 };
bool check(int x, int y) {
    return x >= 0 && x < maxnum && y >= 0 && y < maxnum;
}
int WolfPackDivTwo::calc(vector<int> x, vector<int> y, int m) {

    for (int j = 0; j < maxnum; j++) {
                for (int k = 0; k < maxnum; k++) {
                    res[j][k] =1;
                }
            }
    for (int i = 0; i < x.size(); i++) {
        for (int j = 0; j < maxnum; j++) {
            for (int k = 0; k < maxnum; k++) {
                dp[0][j][k] = dp[1][j][k] = 0;
            }
        }

        dp[0][x[i] + 53][y[i] + 53] = 1;
        for (int t = 1; t < m + 1; t++) {
            int cur = t % 2;
            int pre = (t + 1) % 2;
            for (int j = x[i] + 53 - m - 1; j < x[i] + 53 + m + 1; j++) {
                for (int k = y[i] + 53 - m - 1; k < y[i] + 53 + m + 1; k++) {
                    dp[cur][j][k] = 0;
                    for (int d = 0; d < 4; d++) {
                        int tx = j + dx[d];
                        int ty = k + dy[d];
                        if (check(tx, ty)) {
                            int t = dp[cur][j][k] + dp[pre][tx][ty];
                            dp[cur][j][k] = t % MOD;
                        }
                    }
                }

            }
        }
        int tk = m % 2;
        for (int j = 0; j < maxnum; j++) {
            for (int k = 0; k < maxnum; k++) {
                ll t = (ll) res[j][k] * (ll) dp[tk][j][k];
                res[j][k] = t % MOD;
            }
        }
    }

    int ans = 0;

    for (int i = 0; i < maxnum; i++) {
        for (int j = 0; j < maxnum; j++) {
            ans += res[i][j];
            ans = ans % MOD;

        }
    }
    int resk = ans % MOD;
    return resk;

}